ISC Class 12 Physics: Kirchhoff's Laws with Solved Examples

Why Kirchhoff's Laws Matter in Real Life

Ever wondered how the lights in your house stay on even when you plug in a new gadget? Kirchhoff's laws are the secret sauce that engineers use to keep electricity flowing smoothly.

💡 In Simple Words: Kirchhoff's laws tell us that the amount of electric charge that comes into a junction equals the amount that leaves, and that the total energy gained and lost around any closed loop adds up to zero. Think of water flowing in pipes – what goes in must come out, and the pressure changes around a loop balance out.

Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law (KCL) says that at any node (a point where two or more conductors meet) the sum of currents entering the node equals the sum of currents leaving it. In plain words, charge doesn’t disappear or appear out of nowhere.

Why does this matter? It lets you write equations for complex circuits by just looking at each junction.

Worked Example 1 – Finding an Unknown Current

Consider a simple circuit with three branches meeting at point A. The currents are I₁ = 2 A entering, I₂ = 3 A entering, and I₃ leaving (unknown). Apply KCL:

• Sum of currents entering = Sum of currents leaving
• 2 A + 3 A = I₃
• I₃ = 5 A

That’s it – the unknown current is 5 A leaving the node.

Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law (KVL) states that the algebraic sum of all potential differences (voltage drops) around any closed loop is zero. Imagine walking around a hill: you go up (gain energy) and down (lose energy). By the time you return to the start, the net change is zero.

Potential difference is just another name for voltage – the energy per charge that pushes electrons through a component.

Worked Example 2 – Solving a Loop

Take a loop with a 12 V battery, a 4 Ω resistor (R₁), and a 6 Ω resistor (R₂) in series. Let the current be I (same through both because they’re in series). Write KVL:

• +12 V (battery) – I·4 Ω – I·6 Ω = 0
• 12 – 10I = 0
• I = 1.2 A

Now you can find the voltage drop across each resistor: V₁ = I·R₁ = 1.2 A × 4 Ω = 4.8 V, V₂ = 1.2 A × 6 Ω = 7.2 V. Notice they add up to 12 V, satisfying KVL.

Quick Comparison: KCL vs. KVL

Step‑by‑Step Guide to Solving a Circuit with Both Laws

Follow these steps and you’ll never get stuck on a tricky circuit again.

Key Takeaways

• KCL: total current into a node equals total current out.
• KVL: sum of voltage gains and drops around any closed path is zero.
• Use KCL for junctions, KVL for loops.
• Write equations, solve simultaneously – usually with two or three unknowns.
• Always double‑check by plugging your answers back into the original equations.

📝 Likely Exam Questions

1. Question: In the circuit shown, three currents meet at a node: I₁ = 4 A entering, I₂ = 2 A leaving, I₃ unknown. Find I₃.
   Answer: Using KCL, 4 A = 2 A + I₃ → I₃ = 2 A (leaving).
2. Question: A loop contains a 9 V battery and two resistors, 3 Ω and 6 Ω, in series. Calculate the current.
   Answer: KVL: 9 – I·3 – I·6 = 0 → 9 – 9I = 0 → I = 1 A.
3. Question: For the circuit below, apply KCL at node B and KVL around the left loop to find the unknown current I. (Values: 5 Ω resistor, 10 V source, etc.)
   Answer: KCL: I₁ = I + I₂. KVL left loop: 10 – I·5 = 0 → I = 2 A. Substitute back to get other currents.
4. Question: Explain why KVL must hold even when a battery has internal resistance.
   Answer: The internal resistance adds an extra voltage drop, but the sum of all drops (including that) still equals the emf, keeping the total zero.
5. Question: A circuit has two parallel branches. Branch 1: 2 Ω resistor, Branch 2: 3 Ω resistor. A 12 V battery powers them. Find the current through each branch using KCL and KVL.
   Answer: Voltage across each branch is 12 V (KVL). I₁ = 12/2 = 6 A, I₂ = 12/3 = 4 A. Total current entering the node = 10 A (KCL).